What is the tangent of y=(1+5x)^9,(0,1) ?

The tangent of y=(1+5x)^9,(0,1) is y=45x+1

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tangent of y=(1+5x)^9,(0,1)

tangent of

y = (1 + 5 x)^{9}, (0, 1)

y = 45 x + 1

Solution steps

Find the slope of

y = (1 + 5 x)^{9} : \frac{d y}{d x} = 45 (1 + 5 x)^{8}

Compute the slope of

y = (1 + 5 x)^{9}

(0, 1) : m = 45

Find the line with slope m=

45

and passing through

(0, 1) : y = 45 x + 1

y = 45 x + 1

\int \frac{1}{( x + 4 ) ^{2}} d x

x \to 8 lim (\frac{x}{x - 8})

\frac{d y}{d x} y = (x^{2} + 1)^{3}

y = 3 + 5 x^{2} - 2 x^{3}

\frac{\partial}{\partial x} ((x - 2)^{2} + (5 - x - y)^{2})

What is the tangent of y=(1+5x)^9,(0,1) ?
The tangent of y=(1+5x)^9,(0,1) is y=45x+1